# # 获取能够被3整除的1-10的数值
#
# # list1 = [i for i in range(1,11) if i % 2 == 0]
# # print(list1)
#
# list1 = []
#
# for i in range(0,11):
#     if i % 2 == 0:
#         list1.append(i)
#
#
# #print(f'list集合是:{list1}')
#
#
# tuple1 = (1,2)
# #print(f'元组的集合是:{tuple1}')
#
# list1[0] = 11
# #print(list1)
#
#
# listone = []
# listtwo = []
# for i in range(0,10):
#     for j in range(0,10):
#         listone.append((i,j))
# print(listone)
#
# listtwo = [(i,j) for i in range(0,10) if i % 2 == 0 for j in range(0,10)]
#
#
# print(listtwo)
#
# map1 = {n:n*n for n in range(0,6)}
# map2 = {n:n*n for n in range(6,10)}
#
# print(type(map1.values()))
#
#
# list3 = [1,2,3,7,8]
# list4 = [4,5,6]
# # 两个列表合并为字典   只能统计小的
# dict1 = {list3[i]:list4[i] for i in range(len(list4))}
# print(dict1)
#
#
#
#
#




# # 提取字典中的目标数据
# counts = {'MBP':268,'HP':125,'DELL':201,'Lenovo':199,'acer':99}
# # 需求  提取上述电脑数量大于等于200的字典数据
# count1 = {key:value for (key,value) in counts.items() if value >= 200}
# print(count1)

# 给定一个列表  将列表变为集合 并且集合的值为列表中的数据平方
list1 = [1,2,3,1,2]
set1 = {i ** 2 for i in list1}
print(set1)